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Correct letter and envelope

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Correct letter and envelope

by binaras » Thu Apr 23, 2015 8:39 pm
Hi

need help solving the below

Tanya prepared 4 different letters to be sent to the 4 different addresses.
For each letter she prepared an envelope with it's correct address.
If the four letters are to be put into 4 envelopes at random what is the probability that only one letter will be put in the envelope with its correct address?

Thanks, Binara
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by GMATGuruNY » Fri Apr 24, 2015 12:45 am
Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its complete address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address ?
A. 1/24
B. 1/8
C. 1/4
D. 1/3
E. 3/8
Let the 4 letters be A, B, C and D.
Total ways to arrange the 4 letters = 4! = 24.
Let the correct ordering of the 4 letters be ABCD.

Write out the ways that ONLY A can be put in the correct position:
ACDB
ADBC
Total ways = 2.

Using the same reasoning, there will be 2 ways that ONLY B can be put in the correct position, 2 ways that ONLY C can be put in the correct position, and 2 ways that ONLY D can be put in the correct position.
Thus, the total number of ways to put EXACTLY 1 letter in the correct position = 2+2+2+2 = 8.

Thus:
P(exactly 1 letter is put in the correct position) = 8/24 = 1/3.

The correct answer is D.[/quote]
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by ceilidh.erickson » Sat Apr 25, 2015 12:50 pm
Here's another solution (similar principle, slightly different setup):

Calculate the probability that the first letter goes in the correct envelope: 1/4 chance.
Then, find the probability that the remaining 3 go in the wrong envelopes. For the 2nd letter, there's are 2 wrong envelopes left out of the remaining 3: 2/3
For the 3rd, 1 wrong envelope remaining out of 2: 1/2
For the 4th, there's 1 wrong envelope left: 1/1

So, multiply those together: (1/4)(2/3)(1/2)(1) = 1/12

There's a 1/12 chance of getting the 1st one right and the other 3 wrong. We can infer that there would also be a 1/12 chance of getting the 2nd one right, but the other 3 wrong, and so on. So, it's 1/12 chance for each slot, times 4 slots:
(1/12)(4) = 1/3

The answer is D.
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by mainbhidhruv » Sat Apr 25, 2015 11:59 pm
Dear Experts,

Please let me know if my solution is flawed

P( only 1 letter in correct Envelope) = P ( Letter in Correct Envelope) * P ( IInd letter in incorrect Envelope) * P (IIIrd letter in incorrect Envelope)* P ( 4th letter in the last envelope)
= 1 * 2/3 * 1/2 * 1 = 1/3
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by Aman verma » Sun Apr 26, 2015 2:43 am
binaras wrote:Hi

need help solving the below

Tanya prepared 4 different letters to be sent to the 4 different addresses.
For each letter she prepared an envelope with it's correct address.
If the four letters are to be put into 4 envelopes at random what is the probability that only one letter will be put in the envelope with its correct address?

Thanks, Binara
Hello binaras,

First thing to observe is that the Total Probability consists of symmetrical events. This can shorten your calculation. As Celidh has calculated the probability of the first letter to the correct envelope: 1/12 , you can simply multiply it with the 4 symmetrical events to get 4 X 1/12 = 1/3. Regarding, mainbhidhruv's, it's a bit of an ingenious approach:

P( Letter in Correct Envelope)=certain event,hence, 1,

P ( IInd letter in incorrect Envelope)=2 slots out of remaining 3 slots = 2/3

P (IIIrd letter in incorrect Envelope)=1 slot out of remaining 2 slots = 1/2

P ( 4th letter in the last envelope)= certain event, as only one incorrect slot remains = 1

But, this is a very irregular approach. Students might not remember this approach under exam pressure or may make some faulty calculation. Nevertheless, the solution is correct.
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